document.write( "Question 1178771: solve tan 2x + sec 2x = 7 over the interval [0,2pi) \n" ); document.write( "

Algebra.Com's Answer #808204 by greenestamps(13206) $\"\"$ $\"About$

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\n" ); document.write( " $\"tan%282x%29%2Bsec%282x%29+=+7\"$
\n" ); document.write( " $\"sin%282x%29%2Fcos%282x%29%2B1%2Fcos%282x%29+=+7\"$
\n" ); document.write( " $\"%28sin%282x%29%2B1%29%2Fcos%282x%29+=+7\"$
\n" ); document.write( " $\"sin%282x%29%2B1+=+7cos%282x%29\"$
\n" ); document.write( " $\"7cos%282x%29-sin%282x%29+=+1\"$

\n" ); document.write( "Square both sides and simplify using sin^2(A)+cos^2(A)=1

\n" ); document.write( " $\"49%28cos%282x%29%29%5E2-14cos%282x%29sin%282x%29%2B%28sin%282x%29%29%5E2+=+1\"$
\n" ); document.write( " $\"48%28cos%282x%29%29%5E2-14cos%282x%29sin%282x%29+=+0\"$
\n" ); document.write( " $\"cos%282x%29%2848cos%282x%29-14sin%282x%29%29+=+0\"$
\n" ); document.write( " $\"cos%282x%29=0\"$ or $\"48cos%282x%29-14sin%282x%29=0\"$

\n" ); document.write( "cos(2x)=0 is an extraneous solution; it does not satisfy the original equation (sec(2x) and tan(2x) are both undefined when cos(2x)=0). So

\n" ); document.write( " $\"48cos%282x%29-14sin%282x%29=0\"$
\n" ); document.write( " $\"48cos%282x%29=14sin%282x%29\"$
\n" ); document.write( " $\"48%2F14+=+sin%282x%29%2Fcos%282x%29+=+tan%282x%29\"$

\n" ); document.write( "First solution:
\n" ); document.write( "2x = arctan(48/14) = arctan(24/7) = 1.287 radians (to a few decimal places);
\n" ); document.write( "x = 1.287/2 = 0.6435

\n" ); document.write( "The period of the function is the period of cosine(A), which is 2pi. In the prescribed interval for x, [0,2pi), 2x takes on the values 0 to 4pi. So there is a second solution on the prescribed interval.

\n" ); document.write( "Second solution:
\n" ); document.write( "2x = arctan(48/14)+2pi
\n" ); document.write( "x = 0.6435+pi = 3.7851

\n" ); document.write( " \n" ); document.write( "

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